Dear Candidate,

You can search the same on various portals on internet for step by step method & better understanding. Kindly know, we are limited by our answering options, still I will try to help you.

1. Force on spring when pressure in reservoir is 1N/mm2

P1 = 491N

2. Force on spring when pressure in reservoir is 1.2N/mm2

P2 = 589N

Spring index, C=6, Wahl's factor

K = 1.2525

From tau find the nearest standard diameter, (Coil dia = 6*d = 29.262)

For number of turn, formula for deflection will be used

Delta = [(8*P*D^3)/(G*d^4)]*n , n= effective number of turns

Now,

P2-P1 = 98N

n = 9.2 (after calculating delta)

Actual number of turns with bent = 9.2+2 = 11.2

Now,

Initial deflection at P1 = (491/98)*4 = 20.04mm

Maximum deflection at P2 = (589/98)*4 = 24.04mm

Difference is of 4mm

Now,

Free length of spring

= n'*d + Delta(max) + (n'-1)*1.0

n'= actual number of turns

Therefore free Length of spring,

= 88.86mm(after calculating the whole equation)

Thanks. Hope it helped you a bit.